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I have the following problem: Suppose we have a nonnegative diagonal matrix $A \in \mathbb{R}_+^{n\times n}$ and a matrix $B \in \mathbb{R}^{n\times n}$ with $$Re(\lambda_i(B)) \leq 0,\; i=1,\dots,n$$

where $\lambda_i(B)$ denotes the $i$-th eigenvalue of $B$ and $Re(\cdot)$ denotes the real part.

Is it possible to show that $Re(\lambda_i(AB))\leq 0$ for $i=1,\dots,n$?

I found similar questions but with different conditions on $A$ and $B$:

Any help is appreciated!

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A friend of mine found a simple counterexample:

$A=\begin{bmatrix}4 & 0\\0 & 1\end{bmatrix}, B=\begin{bmatrix}1 & -2\\2 & -2\end{bmatrix}$

For $B$ we have $Re(\lambda_{1}(B)) = Re(\lambda_{2}(B)) =-\frac{1}{2}$ and for $AB = \begin{bmatrix}4 & -8\\2 & -2\end{bmatrix}$ it is $Re(\lambda_{1}(AB)) = Re(\lambda_{2}(AB)) = 1$.

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  • $\begingroup$ Do you mind $\lambda_{1,2}$? $\endgroup$ – user376343 Apr 4 at 8:23
  • $\begingroup$ Yes, that is what i meant. I edited the answer to make it clearer. $\endgroup$ – Paul S Apr 4 at 8:47

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